The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27°C to 42°C. Its energy of activation in J/mol is __________ (Take ln 5 = 1.6094; R = 8.314 J mol$$^{-1}$$)

We are told that when the temperature is raised from 27 °C to 42 °C, the number of molecules having energy greater than the threshold (activation) energy becomes five times. In kinetic theory, the fraction of molecules possessing energy equal to or greater than the activation energy $$E_a$$ at temperature $$T$$ is proportional to the Boltzmann factor $$e^{-E_a/RT}$$. Therefore, for two temperatures $$T_1$$ and $$T_2$$ we can write

$$\frac{\text{Number of molecules at }T_2}{\text{Number of molecules at }T_1} =\frac{e^{-E_a/RT_2}}{e^{-E_a/RT_1}} = e^{-E_a/RT_2 + E_a/RT_1} = e^{E_a\left(\dfrac{1}{RT_1}-\dfrac{1}{RT_2}\right)}.$$

According to the question this ratio equals 5, so

$$5 = e^{E_a\left(\dfrac{1}{RT_1}-\dfrac{1}{RT_2}\right)}.$$

Taking natural logarithm on both sides,

$$\ln 5 = E_a\left(\dfrac{1}{RT_1}-\dfrac{1}{RT_2}\right).$$

Now we insert the numerical data. First convert the given Celsius temperatures to Kelvin:

$$T_1 = 27^\circ\text{C} + 273 = 300\ \text{K},$$

$$T_2 = 42^\circ\text{C} + 273 = 315\ \text{K}.$$

Compute the term in parentheses:

$$\frac{1}{T_1}-\frac{1}{T_2} =\frac{1}{300}-\frac{1}{315} =\frac{315-300}{300 \times 315} =\frac{15}{94500} =\frac{1}{6300}.$$

Substituting $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$ and $$\ln 5 = 1.6094$$ into the logarithmic equation, we get

$$1.6094 = E_a \left(\frac{1}{8.314}\right)\left(\frac{1}{6300}\right).$$

Rearranging for $$E_a$$,

$$E_a = \frac{1.6094 \times 8.314}{\dfrac{1}{6300}} = 1.6094 \times 8.314 \times 6300.$$

First multiply $$1.6094$$ and $$8.314$$:

$$1.6094 \times 8.314 = 13.3805516.$$

Now multiply this result by 6300:

$$13.3805516 \times 6300 = 84\,297.47508.$$

Rounding to the nearest whole number (since the data are given to four significant figures),

$$E_a \approx 84\,297\ \text{J mol}^{-1}.$$

Hence, the correct answer is Option C.